Nonlinear relation between coordinate time and proper time

[craigthone]

For Schwarzschild geomery
$$ds^2=-(1-\frac{2GM}{r})dt^2+(1-\frac{2GM}{r})^{-1}dr^2+r^2d\Omega^2$$
For a Schwarzschild observer , the proper time and coordinate time are related by
$$d\tau=(1-\frac{2GM}{r})^{1/2}dt$$
There is a often used relation between proper time and coordinate time
$$d\tau \sim \exp(-\frac{t}{4GM}) dt$$
I do not know how to relate the two relation, i.e. representing position $r$ in terms of coordinate time $t$. Are there any suggestions?

 

[PeterDonis]

There is a often used relation between proper time and coordinate time

Where are you getting this relation from?

 

[craigthone]

Where are you getting this relation from?

In this paper https://arxiv.org/abs/1505.08108, formula (1.4).
This is the redshift of signals of the infalling observer through the horizon. I know a messy way to derive this. But I want to know if there is some simple way to do it since the relation is so simple. Maybe it is a exercise of some textbook, but I do not remember it.

 

[PeterDonis]

In this paper https://arxiv.org/abs/1505.08108, formula (1.4).

This is not a relation between proper time and coordinate time. It's a relation between the time at infinity $t$ since the instant you dropped something towards a black hole, and the frequency of light you receive at infinity from that object.

 

[PeterDonis]

This is the redshift of signals of the infalling observer through the horizon.

It's the dependence of the redshift of signals you receive at infinity on the Schwarzschild coordinate time $t$ (which is the same as the observer's time at infinity) since you dropped the object. It's not the redshift of a single signal. Nor is it the relationship between proper time and coordinate time for any Schwarzschild observer, which is what you appeared to be comparing it to in the OP.

In short, the paper you linked to is talking about something completely different from the relationship between proper time and coordinate time for a Schwarzschild observer, which is what this thread is about (per your own thread title). So if you want to talk about what's in the paper you linked to, you should start a separate thread on that separate topic.

 

[craigthone]

This is not a relation between proper time and coordinate time. It's a relation between the time at infinity $t$ since the instant you dropped something towards a black hole, and the frequency of light you receive at infinity from that object.

Thanks! I know my mistake now. It is the relation about proper time of signals and coordinate time of signals. Is there any simple way to get it?

 

[PeterDonis]

It is the relation about proper time of signals and coordinate time of signals.

No, it isn't. You are misunderstanding the entire topic of the paper you linked to; it has nothing whatever to do with the proper time vs. coordinate time of anything.

 

[craigthone]

No, it isn't. You are misunderstanding the entire topic of the paper you linked to; it has nothing whatever to do with the proper time vs. coordinate time of anything.

$d\tau$ is proper time of signals.
$dt$ is the time meaured at infinity. You mean I can not call it coordinate time of signals? why?

 

[PeterDonis]

$d\tau$ is proper time of signals.

No, it isn't. The signals are light signals and have no proper time. To the extent it's the proper time of anything, it's the proper time of the infalling object that is emitting the signal; $d\tau$ is the proper time interval between two successive wave crests of the signal, along the infalling object's worldline, when the signal is emitted.

$dt$ is the time meaured at infinity.

The time measured at infinity between two successive wave crests when the signal is received at infinity.

You mean I can not call it coordinate time of signals?

Not in any meaningful sense. The expression $d\tau / dt$ is just the reciprocal of the frequency of the signal when it's received at infinity.

 

[craigthone]

Thanks for your correction

 

[craigthone]

No, it isn't. The signals are light signals and have no proper time. To the extent it's the proper time of anything, it's the proper time of the infalling object that is emitting the signal; $d\tau$ is the proper time interval between two successive wave crests of the signal, along the infalling object's worldline, when the signal is emitted.
The time measured at infinity between two successive wave crests when the signal is received at infinity.
Not in any meaningful sense. The expression $d\tau / dt$ is just the reciprocal of the frequency of the signal when it's received at infinity.

I do want to know if there is some simple way to derive the relation.

 

[PeterDonis]

I do want to know if there is some simple way to derive the relation.

I don't know. The paper just states it without proof, and doesn't give any reference.

 

[sweet springs]

I find the trajectory of bodies falling in BH
t=-2m log(r-2m) + const where m=GM in section 19 of General Theory of Relativity, PAM Dirac. I hope this will help you.

PDF is available at http://amarketplaceofideas.com/wp-c...20-%20General%20Theory%20Of%20Relativity1.pdf

 

[PeterDonis]

I find the trajectory of bodies falling in BH

This gives the Schwarzschild coordinate time along the body's trajectory as a function of radial coordinate $r$, assuming that the body is released from infinity at $t = - \infty$.

 

[sweet springs]

assuming that the body is released from infinity at t=−∞t = - \infty.

No, I am afraid. Dirac there does an approximation that r = 2m + $\epsilon$ where more than second order of $\epsilon$ is neglected. This relation is applicable just near above EH.

 

[PeterDonis]

Dirac there does an approximation that r = 2m + ϵ where more than second order of ϵ is neglected.

Ah, you're right. I was confusing what you quoted with a different formula.

 

[craigthone]

I find the trajectory of bodies falling in BH
t=-2m log(r-2m) + const where m=GM in section 19 of General Theory of Relativity, PAM Dirac. I hope this will help you.

PDF is available at http://amarketplaceofideas.com/wp-content/uploads/2014/08/P%20A%20M%20Dirac%20-%20General%20Theory%20Of%20Relativity1.pdf

I think this works.
For the radial infalling observer near the horizon, $r \rightarrow 2GM$ in the Schwarzschild coordinate, so $dr \approx 0$ and we have
$$d\tau \approx (1-\frac{2GM}{r})^{1/2}dt$$
While from formula of Dirac
$$\frac{r}{2GM}-1 \propto \exp(-t/2GM)$$
Combining the above two formulas, we have
$$ d\tau \propto \exp(-t/4GM) dt $$

 

[PeterDonis]

I think this works.
For the radial infalling observer near the horizon

This can't be all there is to it, because the formula in the paper you linked to is not talking just about phenomena near the horizon. It's talking about the frequency of signals received at infinity. Dirac's discussion says nothing about what happens to outgoing light signals after they are emitted from the infalling object; but that has to be included in any derivation of the formula in the paper you linked to.

In short, you can't just wave your hands and look for formulas with exponentials in them and say that's a derivation.

$r \rightarrow 2GM$ in the Schwarzschild coordinate, so $dr \approx 0$

This is not correct. The infalling object does not stop at $r = 2GM$, and $dr / d\tau$ does not vanish there.

 

[craigthone]

This can't be all there is to it, because the formula in the paper you linked to is not talking just about phenomena near the horizon. It's talking about the frequency of signals received at infinity. Dirac's discussion says nothing about what happens to outgoing light signals after they are emitted from the infalling object; but that has to be included in any derivation of the formula in the paper you linked to.

In short, you can't just wave your hands and look for formulas with exponentials in them and say that's a derivation.
This is not correct. The infalling object does not stop at $r = 2GM$, and $dr / d\tau$ does not vanish there.

For the distant observer, he has $t\rightarrow \infty$, and $r\rightarrow 2GM$. I think $dr=0$ is OK.

 

[PeterDonis]

For the distant observer, he has $t\rightarrow \infty$, and $r\rightarrow 2GM$.

Huh? The distant observer is at $r = \infty$.

I think $dr=0$ is OK.

No, it isn't, because you claimed to be deriving the expression

$$
d\tau \approx \left( 1 - \frac{2GM}{r} \right)^{\frac{1}{2}} dt
$$

for the infalling observer. In order for that to be correct, it would have to be the case that $dr / d\tau \rightarrow 0$ as $r \rightarrow 2GM$. But it isn't. The fact that $dr / dt \rightarrow 0$ as $r \rightarrow 2GM$ is irrelevant.

 

[sweet springs]

Dirac

t=-2m\ log(r-2m) + const

const is explisitly written as

\frac{exp(-t/2m)}{exp(-t_0/2m)}=\frac{1-2m/r}{1-2m/r_0}
where initial position of the falling body m/r_0 <<1 at initial time t=t_0

d\tau=(1-m/r_0)\frac{exp(-t/4m)}{exp(-t_0/4m)}dt

 

[PeterDonis]

Dirac

Is not relevant for the calculation we are trying to do in this thread.

 

[craigthone]

Huh? The distant observer is at $r = \infty$.
No, it isn't, because you claimed to be deriving the expression

$$
d\tau \approx \left( 1 - \frac{2GM}{r} \right)^{\frac{1}{2}} dt
$$

for the infalling observer. In order for that to be correct, it would have to be the case that $dr / d\tau \rightarrow 0$ as $r \rightarrow 2GM$. But it isn't. The fact that $dr / dt \rightarrow 0$ as $r \rightarrow 2GM$ is irrelevant.

Thanks for your corrections again.
How about the following argument, for the infalling observer near the horizon we have
$$\frac{dt}{dr}\approx -\frac{2GM}{r-2GM}=-\frac{2GM}{\varepsilon}$$
This is the formula from Dirac (page 33). Then the proper time for the infalling observer
$$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1} dr^2$$
Then from the $dt$ and $dr$ relation, we know that the $dr^2$ term in the proper time can be neglected compared with the $dt^2$ term.

 

[PeterDonis]

from the $dt$ and $dr$ relation, we know that the $dr^2$ term in the proper time can be neglected compared with the dt2dt2dt^2 term.

No, we don't. Look at the coefficient of the $dr^2$ term; it goes to infinity as $r \rightarrow 2GM$.

Or, more rigorously, plug the following into the metric:

$$
dr^2 = \frac{ \left( r - 2GM \right)^2 }{ \left( 2GM \right)^2 } dt^2
$$

You get:

$$
d\tau^2 = - \frac{r - 2GM}{r} dt^2 + \frac{r}{r - 2GM} \frac{ \left( r - 2GM \right)^2 }{ \left( 2GM \right)^2 } dt^2
$$

Once you've canceled common factors in the second term, both terms go like $r - 2GM$, so both are of similar magnitude.

 

[craigthone]

No, we don't. Look at the coefficient of the $dr^2$ term; it goes to infinity as $r \rightarrow 2GM$.

Or, more rigorously, plug the following into the metric:

$$
dr^2 = \frac{ \left( r - 2GM \right)^2 }{ \left( 2GM \right)^2 } dt^2
$$

You get:

$$
d\tau^2 = - \frac{r - 2GM}{r} dt^2 + \frac{r}{r - 2GM} \frac{ \left( r - 2GM \right)^2 }{ \left( 2GM \right)^2 } dt^2
$$

Once you've canceled common factors in the second term, both terms go like $r - 2GM$, so both are of similar magnitude.

Yes, you are right and you give the answer. Though $dr^2$ terms can not be ignored, it has the same form as $dt^2$ term. In the end we get the relation between $d\tau$ and $dt$. Thanks for all your posts.

 

[craigthone]

$$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1} dr^2$$
$$d\tau^2=\left( 1 - \frac{2GM}{r} \right) dt^2-\left( 1 - \frac{2GM}{r} \right)^{-1}\left(\frac{r-2GM}{2GM}\right)^2dt^2\approx 2\left(1-\frac{2GM}{r}\right)dt^2$$
$$d\tau\approx \sqrt{2}\left(1-\frac{2GM}{r}\right)^{1/2}dt$$

This is what we want.

 

[PeterDonis]

This is what we want.

No, it isn't. Check your algebra. The answer you should be getting is

$$
d\tau^2 = \frac{\left( r - 2GM \right) \left[ r^2 - \left(2GM\right)^2 \right]}{r \left(2GM\right)^2} dt^2
$$

 

[craigthone]

No, it isn't. Check your algebra. The answer you should be getting is

$$
d\tau^2 = \frac{\left( r - 2GM \right) \left[ r^2 - \left(2GM\right)^2 \right]}{r \left(2GM\right)^2} dt^2
$$

You are right again. Why does it not work? This is strange to me.

 

[PeterDonis]

Why does it not work?

Why does what not work?

If you mean, why isn't the final answer just some numerical factor times $\sqrt{1 - 2GM / r}$, the latter is $d\tau / dt$ for a hovering observer--an observer who is at a constant altitude above the horizon. You should not expect $d\tau / dt$ for an infalling observer to have the same dependence on $r$; the infalling observer is moving relative to the hovering observer, and their relative speed is itself $r$ dependent.

 

[sweet springs]

The paper says
----------------
In a black hole, there are two notable exponential behaviors. If a source of fixed frequency is thrown into a black hole, the frequency seen at infinity will be exponentially reshifted,
(1.4)
----------------

I read the initial condition of a frequency source body is at (t_0, r_0) 0< r_0 - GM << 1 close enough to EH. Proper time of the source will be slowed down exponentially approaching to zero in comparison with the world time t.

Even if it was released more far away, it takes only finite proper time to come the initial position above so it does not matter in the discussion of the author who is focusing on Exponential Behaviors.